Electric Field Lines homework

Can anyone help me to understand why equipotential surfaces must always be perpendicular to the electric field lines that pass through them?
My textbook gives what seems to be a simple mathematical explanation, but the logic is escaping me.
It states that dV= -E.ds = 0, (where the "." appears to be dot product of the two vectors -E and ds and so E must be perpendicular to the displacement of the equipotential surface.
I can understand that the dot product=0 means two vectors are perpendicular. I am looking for a more intuitive explanation to really understand what is going on.

Mathematically, is true that a "level surface" for a function is always perpendicular to the gradient.

That is, if F is a function of x,y, z: F(x,y,z), its gradient is the vector Fxi+ Fyj+ Fzk (Fx is the partial derivative with respect to x, etc.) It can be shown that the derivative of F in the direction of any unit vector, v is grad F dot v. Along a "level surface", the function is a constant so taking v to be a vector tangent to that surface, Fv= grad F dot v= 0 since the function is constant on the surface.

Physically, the "potential function" is an anti-derivative of the force function. That is, if F is the potential function and f is the force vector, then f= -grad F. That means that a surface along which F is constant (an "equi-potential" surface) must be perpendicular to the force vector.

Another way of looking at it is this: if you move in the direction of the force vector, your potential goes down (just as moving down in a gravitational field reduces your potential energy). If you move at an angle to the force vector, you will still change your potential depending on the component of your motion parallel to the force vector (If you move down a hill side, you reduce your potential energy by the actual distance down: the component of motion parallel to the gravity force vector). There is no component of motion along a surface parallel to the force vector if and only if the force vector is perpendicular to the surface.